About the equation lgx lg 4 x lg a 2x and discuss the number of solutions

First of all, according to the definition domain of the function, x>0, 4-x>0, a+2x>0

Under the above conditions, the original equation can be rewritten as, x(4-x)=a+2x

x^2-2x+a=0 ..2)

Its δ = 4-4a = 4 (1-a) 0 = > a 1 ...3)

When a=1, there is only one root, x=1, which satisfies the requirement of condition (1) and is therefore the root of the original equation.

When a<1, the two roots of equation (2) are: x1=1+ (1-a) and x2=1- (1-a).

For x1: 0<1+ (1-a)<4...5) ,a+2(1+√(1-a))>0 ..6)

Solution(5) 1-a)<3 => a>-8

Solution (6) a+2(1+ (1-a))>0 => Obviously a>=-2 is constant, and when a<-2, the shift is (1-a)>-a 2+1) =>1-a>a 2 4+a+1 => a(a+8)<0 => a>-8

Therefore, the solution of (6) is: a>-8

i.e. for x1, only if -80 ...8)

Solution (7): 1- (1-a)>0 =>a>0

1- (1-a)<4 = > Heng is established.

Solution (8): a+2(1- (1-a))>0 => 1-a)=-2, a(a+8)>0 =>a>0

That is, for x2, the equation has no solution only if 01 or a<=-8.

The original formula is approved as lg[(x 2+4x-26) (x-3)]=lg10 can be destructible as (x 2+4x-26) (x-3)=10x 2+4x-26=10(x-3).

x^2-6x+4=0

x-3)^2=5

x=3 5

First of all, according to the definition domain of the function, x>0, 4-x>0, a+2x>0

00...1) Under the above conditions, the original equation can be rewritten as, x(4-x)=a+2x

x^2-2x+a=0

Its δ = 4-4a = 4 (1-a).

a≤1...3)

When a=1, there is only one root, x=1

satisfies the requirement of condition (1) and is therefore the root of the original equation.

When a<1, the two roots of equation (2) are: x1=1+ (1-a) and x2=1- (1-a).

For x1: 0<1+ (1-a)<4...

a+2(1+√(1-a))>0

Solution (5)(1-a):<3

a>-8

Solution(6)a+2(1+(1-a))>0

Obviously, a>=-2 is constant, and when a<-2, the shift gives (1-a)>-a 2+1).

1-a>a^2/4+a+1

a(a+8)<0

a>-8

Therefore, the solution of (6) is:

a>-8

i.e. for x1, only if.

Solution (7) 1- (1-a) > 0

a>0

1-√(1-a)<4

Constant establishment. Solution (8) a+2(1- (1-a))>0

(1-a)=-2, a(a+8)>0

a>0

i.e. for x2, only if 01

or a<=-8, the equation has no solution.

First, consider defining the domain: x>0, 4-x>0, and a+2x>0.

1) When a<=-8, the defined domain is an empty set, and the original equation has no real solution.

2) When-80

So there are two solutions to ( ), which are: 1+root(1-a) and 1-root(1-a).

It is easy to determine that 1+ root number(1-a) falls in the defined domain, while 1-root number(1-a) is not in the defined domain.

Thus, the original equation has a real number solution.

3) When a>0, define the domain as 00So there are two solutions to ( ), which are: 1+root(1-a) and 1-root(1-a).

and all fall in the defined domain, where the original equation has two real solutions;

Scenario 2: When a=1, delta=4-4a=0So then ( ) has a solution, i.e., x=1, which is obviously in the domain of definition, and the original equation has a real solution;

Scenario 3: When a>1, delta=4-4a<0. It shows that the original equation has no real solution.

To sum up, the number of real solutions to the original equation is as follows:

1) When a belongs to (-infinity, -8) or (1, + infinity), 0 real solutions;

2) 1 real solution when a belongs to (-8,0) or a=1;

3) When a belongs to (0,1), 2 real solutions.

Of course, this is a primary solution, and if you have learned the derivative, you can use the method of deriving the function to determine the increase or decrease. I won't talk about it here.

Let f(x)=lgx+lg(4-x)-lg(a+2x).

From x>0,4-x>0,a+2x>0:

a<=-8, f(x) defines the domain as ;

At 80, f(x) defines the domain as (0,4).

The monotonicity of f(x) can be judged by using the derivative, and the following conclusions can be made:

1) When a<=-8, the value of x is an empty set, and the number of real solutions is 0;

2) -81, the f(x) images are all below the x-axis, and the number of real solutions is 0

Solution: The original formula is equivalent to lgx(4-x)=lg(a+2x), and x(4-x)=(a+2x), x(4-x)>0, a+2x>0, when a>=0,01, the equation has no solution.

The problem is logarithmic, so there is x>0,4-x>0,a+2x>0 to get x>0,x<4,x>-a 2

lgx+lg(4-x)=lgx(4-x)=lg(a+2x)4x-x²=a+2x

x²-2x+a=0

x=2±√4-4a /2

1± (1-a)/2

This needs to be discussed in the scope of a.

When a>1, there is no solution.

When a<-3, x=1- 1-a 2<1- 1+3 2=0 is rounded off and there is only one root.

When a>-35, x=1+ 1-a 2>1+ 1+35 2=4 is rounded off and there is only one root.

So when 1>a>-3 there are two roots.

lg(a-5x)=lg(9-x^2)-lgx,lg(a-5x)=lg(9-x^2)/x

So a-5x = (9-x 2) x

ax-5x^2=9-x^2

4x^2-ax+9=0

There is a solution, =0

a^2-144=0

a= 12 I don't think I can tease and tear down the teacher, thank you for the pre-harvest dates

(LGA+LGX)(LGA+2LGX)=4Let LGX=Y; lga=t

2y²+3ty+t²-4=0

x>1；then y=lgx>0

So, the roots of the above equations are all positive.

According to Veda's theorem:

y1+y2=-3/2t>0

y1y2=(t²-4)/2>0

Solution: t<-2

So: LGA<-2

0

x>1lgx>10

lg（ax）lg(ax2)=4

lga+lgx)(lga+2lgx)=42lgxlgx+3lgalgx+lgalga-4=0-3lga/2>20

lgalga-4)/2>100

lga<-40/3

lgalga>204

LGA > root number 204 or LGA < - root number 204 = -2 root number 51-2 root number 51 >-40 3

So LGA<-40 3

a<10^(-40/3)

According to the logarithmic arithmic operation, get.

x+1）*（x-2）=4

i.e. x -x-6 = 0

x=3 or x=-2

Test the true number 0 in the original question, and answer x=3